The time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes. Using the 68-95-99.7 rule, if students are given 90 minutes to complete the exam, what percentage of students will not finish?

Here is what they mean by that rule:

68% finish within one std. deviation.
95% finish within two std. deviations (50 to 90 minutes in this case).
99.7% finish withing three std. deviations.

Considering the second group, half (47.5%) will finish within 70 to 90 minutes. Since the fraction above the mean is 50% for a normal distribution, that means 2.5% will not finish at all in 90 minutes.
A more exact value, using a calculator, is 2.3%

To determine the percentage of students who will not finish the exam within the given time of 90 minutes, we need to calculate the proportion of the data that falls above the given time limit.

First, let's find the Z-score corresponding to the value of 90 minutes using the formula:
Z = (X - μ) / σ

Where:
X is the given value (90 minutes)
μ is the mean (70 minutes)
σ is the standard deviation (10 minutes)

Plugging in the values, we have:
Z = (90 - 70) / 10
Z = 20 / 10
Z = 2

The Z-score tells us how many standard deviations above or below the mean the given value is. In this case, 90 minutes is 2 standard deviations above the mean.

Now, we can use the 68-95-99.7 rule (also known as the Empirical Rule) to estimate the proportion of data falling within a certain range for a normal distribution:
- Approximately 68% falls within 1 standard deviation of the mean.
- Approximately 95% falls within 2 standard deviations of the mean.
- Approximately 99.7% falls within 3 standard deviations of the mean.

Since we are interested in the proportion above 2 standard deviations (Z > 2), we can subtract the area under the normal curve for Z ≤ 2 from 1 to get the percentage of students who will not finish the exam.

Using a standard normal distribution table or a calculator, we can find that the area under the curve for Z ≤ 2 is approximately 0.9772.

Now, subtracting from 1, we get:
Percentage = 1 - 0.9772 = 0.0228

Therefore, approximately 2.28% of students will not finish the exam within the given time of 90 minutes.

To determine the percentage of students who will not finish the exam within 90 minutes, we need to find the percentage of students who take longer than 90 minutes to complete the exam.

Given that the time to complete the exam is normally distributed with a mean of 70 minutes and a standard deviation of 10 minutes, we can use the 68-95-99.7 rule (also known as the empirical rule or 3-sigma rule) to estimate the percentage of students who take longer than 90 minutes.

According to the 68-95-99.7 rule:

- Approximately 68% of data falls within one standard deviation of the mean.
- Approximately 95% of data falls within two standard deviations of the mean.
- Approximately 99.7% of data falls within three standard deviations of the mean.

Let's calculate the z-score first. The z-score formula is:

z = (x - mean) / standard deviation

Here, x = 90 minutes, mean = 70 minutes, and standard deviation = 10 minutes.

z = (90 - 70) / 10 = 2

A z-score of 2 indicates that the value of 90 minutes is two standard deviations above the mean.

Since the z-score is 2, we can conclude that approximately 2.5% of data falls beyond two standard deviations above the mean (in the right tail of the distribution).

However, this tells us the percentage of students who will finish the exam in more than 90 minutes. To find the percentage of students who will not finish within 90 minutes, we need to divide this percentage by 2.

Thus, approximately 2.5% / 2 = 1.25% of students will not finish the exam within 90 minutes.